第二题,做出来了!
1.log 14 7 = lg 7 / lg 14
log 14 5 = lg 5 / lg 14
A = lg 7 / lg 14 = lg 7 / (lg 7 + lg 2)
B = lg 5 / lg 14 = (1 - lg2) / (lg 7 + lg 2)
通过以上两式可得:
lg 7 = A / (1 - A + B)
lg 2 = (1 - A) / (1 - A + B)
log 35 28
= lg 28 / lg 35
= (lg 7 + 2lg 2) / (lg 7 + lg 5)
= (lg 7 + 2lg 2) / (lg 7 + 1 - lg2)
= [A / (1 - A + B) + 2(1 - A) / (1 - A + B)] / [A / (1 - A + B) + 1 - (1 - A) / (1 - A + B)]
= [(2 - A) / (1 - A + B)] / [(A + B) / (1 - A + B)]
= (2 - A) / (A + B)
2.分析：由于集合元素具有无序性,因而存在很多可能性,应抓住集合中的常数来解题.
∵ 0∈A,且xy>0,∴lg(xy)=0,
xy=1,从而A={1,x,0},B={0,|x|,y},
若=1,则x=1,则A={1,1,0}与元素互异性矛盾,
若|x|=1,∵x≠1,∴x=-1,此时,A={1,-1,0},B={1,-1,0}.